Physics 21
Fall, 2004
Solution to HW26
34
P62
A 130
×
astronomical telescope is adjusted for a
relaxed eye when the two lenses are 1.25 m apart. What is
the focal length of each lens?
We must Fnd the focal lengths of the objective (
f
o
=
x
)
and the eyepiece (
f
e
=
y
).
Their ratio is the power, and
their sum is the length (since the eye is relaxed, the image
of the objective is just at the focal length of the eyepiece, so
the eye is focused on the image at inFnity. We work in cm:
x
y
= 130
⇒
x
= 130
y
x
+
y
= 125
⇒
x
= 125
−
y
Equating the two expressions for
x
, we Fnd
130
y
= 125
−
y
⇒
131
y
= 125
⇒
y
=0
.
954
Substituting back, we have
f
e
=0
.
954 cm
,f
o
= 124 cm
35
P8
Suppose a thin piece of glass were placed in front
of the lower slit so that the two waves enter the slits 180
◦
out of phase. Describe in detail the interference pattern on
the screen.
The pattern will have the dark spot at the center, because
the waves with equal path lengths will interfere destructively.
The spacing of successive fringes will be the same as if the
waves were in phase.
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '07
 Hickman
 Physics, Light, Wavelength

Click to edit the document details